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345

So, I decided to try it out. I used Audacity to record ~5 seconds of sound that resulted when I dropped a penny, nickel, dime and quarter onto my table, each 10 times. I then computed the power spectral density of the sound and obtained the following results: I also recorded 5 seconds of me not dropping a coin 10 times to get a background measurement. ...


130

If you have the dimensions and material of an object, you can compute both the mass and the normal vibration modes. Just the mass is not enough - a large paper "coin" will have a different fundamental frequency than a small tungsten sphere. A summary of everything that comes below - the result of several edits, and including a nice interaction with the ...


45

It's obviously not a sharp cut-off, but as a general guide sound waves cannot propagate if their wavelength is equal to or less than the mean free path of the gas molecules. This means that even for arbitrarily low pressures sound will still propagate provided the wavelength is long enough. Possibly this is stretching a point, but even in interstellar gas ...


44

In order to properly understand this without any unnecessary "controversy", let's break the whole process of sound generation and perception into 5 important, but completely separate parts. We'll then proceed to explain each part using a few different examples and pieces of derivative logic: Vibration of the vocal folds Transmission of energy from vocal ...


16

The air pressure inside the (intact) bubble is larger than in the surrounding. This pressure difference is called Laplace pressure and is caused by the surface tension between the soap film and the air. When the bubble pops the compressed air expands, thus creating a pressure wave, which you ultimately hear as the typical popping sound.


16

Zero decibels isn't soundless. The decibel scale is a logarithmic one. For sound each 20 decibel step changes the air pressure associated with the sound changes by a factor of ten. So if you take 20dB as a reference, 0dB is a factor of ten quieter and -20dB is a factor of 100 quieter. Completely soundless would be $-\infty$dB. Zero decibels corresponds to ...


15

This is not an advertisement. Under the rubric of "do try this at home", I wanted to share one more thing that I discovered after writing my previous answer - but it is so unrelated to that answer that I thought it better to write this as a separate post. I discovered two interesting things. First, when you spin a coin on a hard surface, it "rings" with ...


11

I expect that there is no minimal pressure. A sound wave is a density wave. If the particles are close to each other they will interact due to strong forces like van der Waals force and Coulomb force. Reducing the pressure in constant volume leads to long distances between the particles. Lets assume the particles have a huge distance and we ignore even ...


6

I believe this is happening because the racquetball court is acting like a acoustic waveguide. Reflections off of the walls create a resonance that makes the propagating waves show dispersion so that a sharp localized sound gets stretched out into the "chirp" that you hear. I tried to simulate the sound inside a racketball court: Above you see a slowed ...


4

Have a look at this black body radiation spectrum, which is approximately the spectrum of "light", electromagnetic radiation, that a body radiates because of the intrinsic kinetic degrees of freedom of the molecules. Look at the frequency spectrum for 300K, about room temperature. Acoustic frequencies are of the order of a few thousand Hertz, infrared is ...


3

I don't mean to take anything away from the previous great answers, but the "simple and to the point" answer is, a very qualified, yes. By qualified, I mean one must know the coin's composition, thickness, diameter(or shape), density distribution, country of manufacture, etc. If we make assumptions and restrictions, then it becomes possible to calculate ...


3

It should be obvious that the linear speed is inversely proportional to the distance from the center to that point on that groove. Yes, speed varies over a record, so the wavelength of the wiggles in the groove gets shorter for the same frequency as you get further into the record. However, the master record was created with this same phenomenon in ...


3

With $t\to-it$ and $P(t)=\delta(t-t_0)$ and some constants redefinition, this is an equation for the Green's function of the Schrödinger equation $-i \hbar \partial_t \psi = \left(-\frac{\hbar^2}{2m} \Delta + V\right) \psi$ Where $V \sim cm$ is a constant. In most contexts, one needs either $V=0$ or $V=V(r)$. The solutions however become oscillatory which ...


3

This page quotes the pressure inside a soap bubble as $\frac {4\gamma }R$, where $\gamma$ is the surface tension, about $25\text { dyne}/\text{cm }$ for soapy water, and $R$ is the radius of the bubble. For $R=1$ cm, the pressure is then $100 \text { dyne}/\text{cm}^2 = 10 \text{ Pa}$. This is released when the bubble pops. It doesn't seem like much with ...


3

Effectively zero, but it takes a mental stretch to get there. When you're dealing with a gas, lower pressure means that there is a longer mean free path, meaning the atoms/molecules can be expected to go longer and longer between collisions. You can get this either by spacing out the particles more (lower density) or by slowing them down (lower ...


3

This page gives a chart of Young's modulus over temperature for various metals. Taking the top line of the table, the modulus drops from 31.4 Msi at -325F (-200C) to 24.2 Msi at 800F (427C). Due to thermal expansion, you will have more square inches. Using a linear expansion of $12E-6 K^{-1}$ the area of a bar will increase $1.5\%$ The longitudinal ...


2

Decibels are a unit of measurement expressing a logarithmic ratio between the intensity of sound and a given fixed intensity. When you see a negative value in decibels does not mean of course that you have negative sound: such a concept has no physical meaning. What actually has physical meaning is the power or the intensity of sound and those are never ...


2

Some good points made in different answers. I just want to add my two cents. The short answer is "yes - it can appear louder, and it can be louder". First - appearance. If you have a loud point source far away, and another source closer by, it is possible that the close source clouds the faraway sound. Imagine a faraway train and a nearby radio playing ...


2

By going to the Analyze Menu and selecting Plot Spectrum: You can generate the power spectral density with different windowing functions and settings: As well as export it.


1

It depends on the direction in which you choose the $y$-axis. If a positive $y$ means a displacement to the left, when this figure is accurate. But you right, that choice is at least unconventional and should have been indicated in the figure.



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